Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int CASE = 1;
const int mxn = 1e5 + 1;
long long __gcd(long long a, long long b) { return !b ? a : __gcd(b, a % b); }
bool prime(long long n) {
  if (n <= 1) return false;
  for (int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return false;
  return true;
}
int convert(int n, int base) {
  int ans = 0;
  while (n) {
    ans += n % base;
    n /= base;
  }
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  map<int, int> mp;
  vector<long long> v;
  long long i = 1;
  while (i * (i + 1) / 2 <= n) {
    v.push_back(i * (i + 1) / 2);
    mp[v.back()]++;
    i++;
  }
  for (long long i : v)
    if (mp[n - i]) {
      cout << "YES";
      return;
    }
  cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}